A circuit has a section AB as shown in figure. The emf of the source equals E = 10 V, the capacitor capacitances are equal to C₁ = 1.0 μF and C₂ = 2.0 μF and the potential difference V_A – V_B = 5.0 V. Find the voltage across each capacitor.

Solution

Given, the emf of the source, E=10 V
Capacitor, C₁= 1 μF
Capacitor, C₂= 2 μF
Potential difference between A and B, V_A – V_B = 5.0 V
Let the charge distribution be as shown in figure

Using the formula,

∴

V_{A} - V_{B} = \frac{q}{C_{1}} - E + \frac{q}{C_{2}}

(V_{A} - V_{B}) + E = q (\frac{1}{C_{1}} + \frac{1}{C_{2}})

= \frac{q (C_{2} + C_{1})}{C_{1} C_{2}}

∴

q = \frac{[(V_{A} - V_{B}) + E] C_{1} C_{2}}{C_{1} + C_{2}}

Voltage across

C_{1}

V_{1}

\frac{q}{C_{1}} = \frac{[(V_{A} - V_{B}) + E] C_{2}}{C_{1} + C_{2}}

= \frac{(5 + 10) 2.0}{1.0 + 2.0} = 10 volt

Voltage across

C_{2}

, V₂ =

\frac{q}{C_{2}} = \frac{[(V_{A} - V_{B}) + E] C_{1}}{C_{1} + C_{2}}

= \frac{(5 + 10) 1.0}{1.0 \times 2.0} = 5 volt

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A circuit has a section AB as shown in figure. The emf of the source equals E = 10 V, the capacitor capacitances are equal to C₁ = 1.0 μF and C₂ = 2.0 μF and the potential difference V_A – V_B = 5.0 V. Find the voltage across each capacitor.